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Use equation [tex]I=\int r^2dm[/tex] to calculate the moment of inertia of a uniform, solid disk with mass M and radius R for an axis perpendicular to the plane of the disk and passing through its center.

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Kazeemsodikisola

Answer:

Explanation:

Given that

I= ∫r²dm

We need to integrate over the volume of the disc, which is not smart enough, so we brake the disc into rings and integrate from the center r=0 to r=R, as shown in attachment

Since we know the moment of inertial of a ring to be

I=mr²

We cannot integrate the r² with respect to m, so we are going to write dm as a function of r

dm

So we assume that

The ratio of the area of the rings to the whole disc is equal to ratio of their masses.

dm/M=2πrdr/πR²

Therefore,

dm=2rM/R² dr

So we are going to replace dm in the integral

I= ∫r²dm

I= ∫r²(2rM/R²) dr. From 0 to R

Where 2, M and R are constant,

I= 2M/R² ∫r³dr. From r=0 to r=R

I= 2M/R² (r⁴/4) From r=0 to r=R

I=2M/R² (R⁴/4 - 0)

I=2M/R² × R⁴/4

I=MR²/2.

This is the moment of inertial of the disc and it doesn't depend on the thickness of the disc.

Ver imagen Kazeemsodikisola

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